CodeForces 602 A. Two Bases

题目描述：

水题，只是注意精度，还有如果你要用pow()函数来求解幂，那么注意pow()返回值为double，如果你用long long int 那么就wrong了，及时类型转换可是在大数据时还是出现数据错乱，不知道为什么。

思路：都转换成十进制然后比较就行

提后分析：

用c做十进制的输入输出格式
int %d
long int %ld
long long int %I64d
unsigned int %d 
unsigned long int %ld
unsigned long long int %I64d
double 输入%lf
float 输入 %f
输出 %lf或%f都行
long double 就不知道用什么了 还是用cin吧！！！

题目：

After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbersX and Y realised that they have different bases, which complicated their relations.

You're given a number X represented in baseb_x and a numberY represented in base b_y. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and b_x (1 ≤ n ≤ 10,2 ≤ b_x ≤ 40), wheren is the number of digits in the b_x-based representation ofX.

The second line contains n space-separated integersx₁, x₂, ..., x_n (0 ≤ x_i < b_x) — the digits of X. They are given in the order from the most significant digit to the least significant one.

The following two lines describe Y in the same way: the third line contains two space-separated integersm and b_y (1 ≤ m ≤ 10,2 ≤ b_y ≤ 40,b_x ≠ b_y), wherem is the number of digits in the b_y-based representation ofY, and the fourth line contains m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y) — the digits of Y.

There will be no leading zeroes. Both X andY will be positive. All digits of both numbers are given in the standard decimal numeral system.

Output

Output a single character (quotes for clarity):

• '<' if X < Y
• '>' if X > Y
• '=' if X = Y

Sample test(s)

Input

6 21 0 1 1 1 12 104 7

Output

=

Input

3 31 0 22 52 4

Output

<

Input

7 1615 15 4 0 0 7 107 94 8 0 3 1 5 0

Output

>

Note

In the first sample, X = 101111₂ = 47₁₀ = Y.

In the second sample, X = 102₃ = 21₅ andY = 24₅ = 112₃, thusX < Y.

In the third sample, andY = 4803150₉. We may notice thatX starts with much larger digits and b_x is much larger than b_y, so X is clearly larger than Y.

代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string>#include <stack>#include <queue>#include <algorithm>#include <math.h>using namespace std; long double x, ax, y, by; long double DX, DY;//存储转换后十进制的数据 long double toDecimal( long double a, long double ax){     long double temp = 0;     long double d = 0;    for( int i=(int)a; i>0; i--)    {        cin>> temp;        d += temp * pow(ax, i-1);    }    return d;}int main(){    while( cin>> x >> ax )    {        long double x1 = toDecimal(x, ax);        cin>> y >> by;        long double y1 = toDecimal(y, by);        if(x1 > y1)            cout<<">"<<endl;        else            if(x1 == y1)            cout<<"="<<endl;        else            cout<<"<"<<endl;    }    return 0;}